Question

Identify the homoleptic complexes with odd number of d-electrons in the central metal:

(A) [FeO₄]²⁻      (B) [Fe(CN)₆]³⁻

(C) [Fe(CN)₅NO]²⁻      (D) [CoCl₄]²⁻

(E) [Co(H₂O)₃F₃]

Choose the correct answer from the options given below :

• A. (B) and (D) only
• B. (C) and (E) only
• C. (A), (B) and (D) only
• D. (A), (C) and (E) only

Hint:

To identify complexes with odd numbers of d-electrons, determine the oxidation state of the central metal and subtract the oxidation state from the total number of electrons in the neutral atom.

Answer 1

The Correct Option is A

- (A) \([FeO_4]^{2-}\): Iron in this complex is in the +2 oxidation state, with \( d^6 \) electrons. Since 6 is an even number, this complex does not meet the requirement for an odd number of d-electrons.
- (B) \([Fe(CN)_6]^{3-}\): Iron in this complex is in the +3 oxidation state, with \( d^5 \) electrons. This gives an odd number of d-electrons, so this is a homoleptic complex with odd d-electrons.
- (C) \([Fe(CN)_6]^{2-}\): Iron is in the +2 oxidation state with \( d^6 \) electrons. This complex has an even number of d-electrons.
- (D) \([CoCl_4]^{2-}\): Cobalt in this complex is in the +2 oxidation state, with \( d^7 \) electrons. This gives an odd number of d-electrons, so this is a homoleptic complex with odd d-electrons.
- (E) \([Co(H_2O)_6]^{3+}\): Cobalt in the +3 oxidation state has \( d^6 \) electrons, which is an even number.

Thus, the correct answer is (1) (B) and (D) only.

Answer 2

Step 1: Understand the question.
We are asked to identify the homoleptic complexes (complexes having identical ligands) that contain a central metal atom with an odd number of d-electrons.

Step 2: Analyze each complex.
(A) [FeO₄]²⁻
Oxidation state of Fe: Let it be x.
\[ x + 4(-2) = -2 \Rightarrow x = +6 \] Electronic configuration of Fe (Z = 26): [Ar] 3d⁶ 4s²
Fe⁶⁺ → [Ar] 3d⁰ (no d-electrons)
Therefore, it has 0 d-electrons and is not odd. Also, this complex is homoleptic but does not have an odd number of d-electrons.

(B) [Fe(CN)₆]³⁻
Oxidation state of Fe: Let it be x.
\[ x + 6(-1) = -3 \Rightarrow x = +3 \] Fe³⁺ → [Ar] 3d⁵
Thus, there are 5 d-electrons (odd number). The ligands (CN⁻) are identical, making it a homoleptic complex.
Hence, (B) is correct.

(C) [Fe(CN)₅NO]²⁻
This complex contains two different types of ligands (CN⁻ and NO), so it is heteroleptic and does not qualify as homoleptic. Hence, (C) is incorrect.

(D) [CoCl₄]²⁻
Oxidation state of Co: Let it be x.
\[ x + 4(-1) = -2 \Rightarrow x = +2 \] Co²⁺ → [Ar] 3d⁷
Thus, it has 7 d-electrons (odd number). All ligands are identical (Cl⁻), making it homoleptic.
Hence, (D) is correct.

(E) [Co(H₂O)₃F₃]
Contains two different types of ligands (H₂O and F⁻), making it heteroleptic. Hence, (E) is incorrect.

Step 3: Conclusion.
Homoleptic complexes with an odd number of d-electrons are (B) and (D).

Final Answer:
\[ \boxed{(B) \text{ and } (D) \text{ only}} \]